(1)令a=b=0
则f(0)=f(0)+f(0)
∴f（0）=0
令a=1
则f(b)=f(1)+f(b)
∴f(1)=0
(2)
f(36)=f(6×6)=f(6)+f(6)=2f(6)=2f(2×3）=2[f(2)+f(3)]=2(p+q)